Questions: Using the relationship 22.4 Liters = 1 mole and the chemical equation below, calculate the mass of calcium carbonate needed to produce 39 mL of carbon dioxide. CaCO3(s) + 2 H3O+(aq) → Ca^2+(aq) + 3 H2O(l) + CO2(g)

Solution Steps

First, we need to convert the volume of CO\(_2\) gas produced (39 mL) to moles using the relationship that 22.4 liters of gas is equivalent to 1 mole at standard temperature and pressure (STP).

\[ \text{Moles of CO}_2 = \frac{\text{Volume of CO}_2}{22.4 \text{ L}} = \frac{39 \text{ mL}}{22.4 \text{ L}} \times \frac{1 \text{ L}}{1000 \text{ mL}} \]

\[ \text{Moles of CO}_2 = \frac{39}{22400} = 0.0017411 \text{ moles} \]

From the balanced chemical equation, we see that 1 mole of CaCO\(_3\) produces 1 mole of CO\(_2\). Therefore, the moles of CaCO\(_3\) needed are the same as the moles of CO\(_2\) produced.

\[ \text{Moles of CaCO}_3 = 0.0017411 \text{ moles} \]

Next, we convert the moles of CaCO\(_3\) to mass using its molar mass. The molar mass of CaCO\(_3\) (calcium carbonate) is:

\[ \text{Molar mass of CaCO}_3 = 40.08 \text{ (Ca)} + 12.01 \text{ (C)} + 3 \times 16.00 \text{ (O)} = 100.09 \text{ g/mol} \]

\[ \text{Mass of CaCO}_3 = \text{Moles of CaCO}_3 \times \text{Molar mass of CaCO}_3 = 0.0017411 \text{ moles} \times 100.09 \text{ g/mol} \]

\[ \text{Mass of CaCO}_3 = 0.1742 \text{ g} \]

Final Answer